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Proof step 2 from proof int2uint-fill (previous step)

0 ≤ j ⊦ bits2uint(fillsgn(n, int2bits(j))) = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ¬ bits.first ∧ bits ≠ [] → bits2int(bits) = bits2uint(bits)
Proof step 4:
¬(¬ fillsgn(n, int2bits(j)).first ∧ fillsgn(n, int2bits(j)) ≠ []), 0 ≤ j ⊦ bits2uint(fillsgn(n, int2bits(j))) = j
Proof step 7:
¬ fillsgn(n, int2bits(j)).first ∧ fillsgn(n, int2bits(j)) ≠ [], 0 ≤ j ⊦ bits2int(fillsgn(n, int2bits(j))) = j

Rule argument

name: bits2int-pos-02 rotate: T subst: [bits] → [fillsgn(n, int2bits(j))]

Additional information

Used terms: fillsgn(n, int2bits(j))